Morphocompletion for #4683 ⟨a, b | aabbaaab=baa

Solved by morph:2/0,4/0. (See Morphocompletion.)

Step 1

Checking up to 20 rules for overlaps.

Rewriting system is not complete: arenaTotalLimit

#Rule
1. aabbaaab ⇒ baa
2. baabaaab ⇒ aabbabaa
3. aabbaaaaabbabaa ⇒ baaaabaaab
4. baaaabaaabbbaaab ⇒ aabbaaaaabbabbaa
5. baabaaaaabbabaa ⇒ aabbabaaaabaaab
6. baaaabaaababbaaab ⇒ aabbaaaaabbababaa
7. aabbaaaaabbaaaaabbabbaa ⇒ baaaaaabaaabbbaaab
8. baaaaaabaaabbbaaabbbaaab ⇒ aabbaaaaabbaaaaabbabbbaa
9. baaaabaaabbbaaaaabbabaa ⇒ aabbaaaaabbabbaaaabaaab
10. baabaaaaabbaaaaabbabbaa ⇒ aabbabaaaaaabaaabbbaaab
11. aabbaaaaabbaaaaabbababaa ⇒ baaaaaabaaababbaaab
...

Collecting factors up to length 7, frequency 4:

Length 2:[2/0] aa, [2/1] ba, [2/2] ab, [2/3] bb
Length 3:[3/0] aaa, [3/1] baa, [3/2] aab, [3/3] bba
Length 4:[4/0] aaab, [4/1] baaa, [4/2] aaaa, [4/3] aabb
Length 5:[5/0] baaaa, [5/1] aabba, [5/2] baaab, [5/3] aaaab
Length 6:[6/0] baaaaa, [6/1] aabbaa, [6/2] aaaaab, [6/3] aaabba

Considering [length 2 / frequency 0] aa=c.

Step 2

Checking up to 20 rules for overlaps.

Rewriting system is not complete: roundsLimit

#Rule
1. ca ⇒ ac
2. cbbacb ⇒ bc
3. bcbacb ⇒ cbbabc
4. aa ⇒ c
5. cbbaccbbabc ⇒ bccbacb
6. bcbaccbbabc ⇒ cbbabccbacb
7. bccbacbbbacb ⇒ cbbaccbbabbc
8. cbbaccbbabac ⇒ bccbacba
9. cbbaccbbaccbbabbc ⇒ bcccbacbbbacb
10. bcbaccbbabac ⇒ cbbabccbacba
11. bcbaccbbaccbbabbc ⇒ cbbabcccbacbbbacb
12. bccbacbabbacb ⇒ cbbaccbbababc
13. bccbacbbbaccbbabc ⇒ cbbaccbbabbccbacb
14. bcccbacbbbacbbbacb ⇒ cbbaccbbaccbbabbbc
15. cbbaccbbaccbbababc ⇒ bcccbacbabbacb
16. cbbaccbbaccbbabbac ⇒ bcccbacbbbacba
17. bcbaccbbaccbbababc ⇒ cbbabcccbacbabbacb
18. bcbaccbbaccbbabbac ⇒ cbbabcccbacbbbacba
19. bccbacbabbaccbbabc ⇒ cbbaccbbababccbacb
20. cbbaccbbaccbbababac ⇒ bcccbacbabbacba
...

Collecting factors up to length 6, frequency 7:

Length 2:[2/0] cb, [2/1] ba, [2/2] ac, [2/3] bb, [2/4] bc, [2/5] cc, [2/6] ab
Length 3:[3/0] bac, [3/1] cbb, [3/2] bba, [3/3] ccb, [3/4] acc, [3/5] bab, [3/6] acb
Length 4:[4/0] cbba, [4/1] bbac, [4/2] bacc, [4/3] accb, [4/4] ccbb, [4/5] bacb, [4/6] bbab
Length 5:[5/0] cbbac, [5/1] ccbba, [5/2] baccb, [5/3] accbb, [5/4] bbacc, [5/5] cbbab, [5/6] bcbac

Considering [length 4 / frequency 0] cbba=d.

Step 3

Rewriting system is complete. See a, b | aabbaaab=baa.